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Proof step 23 from proof int2int-lastn-negneg (previous step)

lastn(m, int2bits(i)) ≠ [] ∧ lastn(m, int2bits(i)).first, # bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), (bits2int(bits) + 1) * 2 ^ n→i(m) = (i + (~ bits2int(lastn(m, int2bits(i))) - 1)) + 1, bits.first, m ≤ # int2bits(i), lastn(m, int2bits(i)).first, i < 0 ⊦ i + 2 ^ n→i(m) = 0, bits2int(lastn(m, int2bits(i))) = i % 2 ^ n→i(m), m = 0, bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
i = (bits2int(bits) + 1) * 2 ^ n→i(m) + bits2int(lastn(m, int2bits(i))), int2bits(i) = bits + lastn(m, int2bits(i)), # bits = # int2bits(i) - m, i < 0, lastn(m, int2bits(i)).first, m ≤ # int2bits(i), bits.first ⊦ bits = [], m = 0, bits2int(lastn(m, int2bits(i))) = i % 2 ^ n→i(m), i + 2 ^ n→i(m) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ lastn(n, bits) = [] ↔ n = 0;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i - j + i = ~ j;% ⊦ i + ~ j = i - j;% ⊦ ~ i - j = ~(i + j);